Respuesta :
Answer:
42.6 N/m
Explanation:
The spring constant can be found by using Hooke's law:
F = kx
where
F is the force applied
k is the spring constant
x is the stretching/compression of the spring
In this problem, the mass applied is
m = 0.10 kg
so the force applied is the weight of the mass:
[tex]F=mg=(0.10 kg)(9.8 m/s^2)=0.98 N[/tex]
The stretching of the spring is
x = 2.3 cm = 0.023 m
So the spring constant is
[tex]k=\frac{F}{x}=\frac{0.98 N}{0.023 m}=42.6 N/m[/tex]
Answer: The value of spring constant is 42.6 N/m
Explanation:
Force is defined as the mass multiplied by the acceleration of the object.
[tex]F=m\times g[/tex]
where,
F = force exerted on the spring
m = mass = 0.10 kg
g = acceleration due to gravity = [tex]9.8m/s^2[/tex]
Putting values in above equation, we get:
[tex]F=0.10kg\times 9.8m/s^2\\\\F=0.98N[/tex]
To calculate the spring constant, we use the equation:
[tex]F=k\times x[/tex]
where,
F = force exerted on the spring = 0.98 N
k = spring constant = ?
x = length of the spring = 2.3 cm = 0.023 m (Conversion factor: 1 m = 100 cm)
Putting values in equation 1, we get:
[tex]0.98N=k\times 0.023m\\\\k=\frac{0.98N}{0.023m}=42.6N/m[/tex]
Hence, the value of spring constant is 42.6 N/m